Question

Please help asap!!

An agency surveyed people worldwide, asking them, "Do you live in a household with more than one car?" Of the 1399 respondents surveyed in India, 307 said they live in a household with more than one car. In Canada, of the 809 respondents, 584 said they lived in a household with more than one car. Find the lower limit for a 99% confidence interval for the difference in the proportions of households with more than one car between India and Canada ( PI – PC), assuming 20.005 = 2. 58. Note: 1 - Your final answer should be expressed as a decimal and not a percentage. 2- Round any intermediate numbers to 4 decimal places. 3- Round your final answer to 3 decimal places. Enter your final answer with 3 decimal places.

Answer 1

Here, , n1 = 1399 , n2 = 809
p1cap = 0.2194 , p2cap = 0.7219

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.2194 * (1-0.2194)/1399 + 0.7219*(1-0.7219)/809)
SE = 0.0193

For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.2194 - 0.7219 - 2.58*0.0193, 0.2194 - 0.7219 + 2.58*0.0193)
CI = (-0.552 , -0.453)

Lower limit = -0.552